I have to show that $\displaystyle \binom{n}{k}<\binom{n}{l}$ for $\displaystyle 0 \le k <l \le\frac{n}{2}$ where $n,k,l$ are an integers. I think I solved it but I'm not sure if my approach is good and I'd like to know if there is easier way to solve it
My solution:
We can rewrite our inequality as $\displaystyle \frac{1}{(n-k)(n-k-1)\cdot...\cdot (n-l+1)}<\frac{1}{l\cdot (l-1)\cdot...\cdot(k+1)}$ and it's equivalent to
$(n-k)(n-k-1)\cdot...\cdot (n-l+1)>l\cdot (l-1)\cdot...\cdot(k+1)$
both sides have the same number of elemetns since LHS has $n-k-(n-l+1)+1=l-k$ elements and RHS has $l-(k+1)+1=l-k$
what is more we have $\displaystyle n-k>\frac{n}{2}$ what implies $n-k>l$ and $n-l+1>k+1$ so our inequality holds.
Is my solution OK ? If you have other idea how to prove it feel free to share it.
Other idea: induction.
Take the statement: $$\binom{n}{k}<\binom{n}{k+1}\text{ for each } k\in\left\{ 0,\dots,\lfloor\frac{n}{2}\rfloor\right\} $$
as inductionhypothese and prove this to be true when $n$ is replaced by $n+1$.
This on base of the hypothese and the well-known rule that: $$\binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k}$$